Calculating Scattered Photon Energy and Recoil Angle in Compton Effect

In summary, the conversation discusses finding the energy of a 650-keV gamma ray scattered at 110°, the kinetic energy of the scattered electron, and the recoil angle of the electron. The Compton effect equation and the conservation of energy equation are used to solve the problem. It is determined that momentum should be used instead of energy to find the recoil angle.
  • #1
wolski888
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Homework Statement


A 650-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 110°, the kinetic energy of the scattered electron, and the recoil angle of the electron.

Homework Equations


Compton effect equation:
Δλ = [itex]λ^{'}[/itex] - λ = (h/mc) (1-cosθ)

Conservation of Energy:
hf + m[itex]c^{2}[/itex] = h[itex]f^{'}[/itex] + [itex]E_{e}[/itex]

hf is initial photon energy, m[itex]c^{2}[/itex] is electron energy before scattering, h[itex]f^{'}[/itex] is the energy of scattered photon with new frequency f prime, and [itex]E_{e}[/itex] is the energy of the recoil electron with mc^2 and Kinetic Energy.

The Attempt at a Solution


First of all I am confused about the question. Are they saying a gamma ray is scattered? Or does a gamma ray hit the electron producing a scattering photon and the electron recoils?

For now, I am thinking that the gamma ray hits the electron (the latter mentioned).

For the recoil angle, I am thinking that it should be 80° since momentum is conserved and so the electron goes in the opposite direction, so 80°.

To find the energy of scattered photon I need to find the new frequency. My thought is that I need to find the new wavelength so that I can use f=c/λ. I can find the new wavelength using Compton Effect equation shown above. I know θ is 110°, and h,m,c are constants. But what about initial λ? Can I get that from the 650-keV? So E = 650-keV = hc/λ using h in terms of eV I can get λ. Right? Now I am all set to find the new wavelength.With that I can find the new frequency, and then find the scattering energy of the photon.

Now I should have all the information to find the Kinetic Energy using the Conservation of energy equation above.

Am I going the right way, any comments would be appreciated. Thanks! :)
 
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  • #2
First of all I am confused about the question. Are they saying a gamma ray is scattered? Or does a gamma ray hit the electron producing a scattering photon and the electron recoils?
What's the difference? I don't understand the distinction you're trying to make.

For the recoil angle, I am thinking that it should be 80° since momentum is conserved and so the electron goes in the opposite direction, so 80°.
You'll have to rethink this.

To find the energy of scattered photon I need to find the new frequency. My thought is that I need to find the new wavelength so that I can use f=c/λ. I can find the new wavelength using Compton Effect equation shown above. I know θ is 110°, and h,m,c are constants. But what about initial λ? Can I get that from the 650-keV? So E = 650-keV = hc/λ using h in terms of eV I can get λ. Right? Now I am all set to find the new wavelength.With that I can find the new frequency, and then find the scattering energy of the photon.
Good plan. You don't need to find the frequency though. The equation you already mentioned,
##E = hc/\lambda##, allows you to relate energy to wavelength directly.
 
  • #3
Forget about the first question then. What do you mean rethink this? Should I use the Kinetic Energy of the recoiled electron to determine the angle?

EDIT: Or should I use conservation of momentum?
 
  • #4
You need to use momentum. Energy doesn't have a direction whereas momentum does.
 
  • #5
Many thanks! I'll get to it then.
 

Related to Calculating Scattered Photon Energy and Recoil Angle in Compton Effect

1. What is the Compton Effect?

The Compton Effect is a phenomenon in which X-rays or gamma rays, when directed at a material, lose some of their energy and change direction. This occurs due to the scattering of photons off the electrons in the material.

2. Who discovered the Compton Effect?

The Compton Effect was discovered by American physicist Arthur Compton in 1923. He received the Nobel Prize in Physics in 1927 for his work on the Compton Effect.

3. What is the significance of the Compton Effect?

The Compton Effect provided evidence for the dual nature of light, which is both a wave and a particle. It also helped to confirm the concept of energy quantization, which is a fundamental principle in quantum mechanics.

4. How is the Compton Effect used in science and technology?

The Compton Effect is used in various applications such as medical imaging, where it is used to produce images of the body's internal structures. It is also used in spectroscopy to determine the structure of molecules and in particle accelerators to manipulate and measure the energy of particles.

5. What factors affect the magnitude of the Compton Effect?

The magnitude of the Compton Effect is affected by the energy of the incident photon, the angle of scattering, and the atomic number of the material. Higher energy photons, larger scattering angles, and materials with higher atomic numbers will result in a larger Compton Effect.

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